The velocity just before he touches the snow is given by v1
Now,
using equation of motion for his free fall to snow
v12−u2=2a1s1
u=0
a1=9.8sec2m
s1=488ft=148.742m
Hence,
v1= 54secm
Now,
using equation of motion for decceleration by snow
v2−v12=2a2s2
s2=4.5ft=1.3716m
v=0
Solving it we get,
a2=−1063sec2m
his average acceleration as he slowed to a stop is 1063sec2m
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